The Unapologetic Mathematician

Mathematics for the interested outsider

Lifting and Descending Representations

Let’s recall that a group representation is, among other things, a group homomorphism. This has a few consequences.

First of all, we can consider the kernel of a matrix representation X. This is not the kernel we’ve talked about recently, which is the kernel of a G-morphism. This is the kernel of a group homomorphism. In this context, it’s the collection of group elements g\in G so that the image X(g) is the identity transformation. We call this subgroup N\subseteq G. If N is the trivial subgroup, we say that X is a “faithful” representation, since it doesn’t send any two group elements to the same matrix.

Now, basic group theory tells us that N is a normal subgroup, and so we can form the quotient group G/N. I say that the representation X “descends” to a representation of this quotient group. That is, we can define a representation Y by Y(gN)=X(g) for all cosets gN\in G/N. We have to check that this doesn’t depend on which representative g of the coset we choose, but any other one looks like g'=gn for n\in N. Then Y(g')=Y(g)Y(n)=Y(g), since Y(n) is the identity matrix by definition.

I say that Y is a faithful representation. That is, the only coset that Y sends to the identity matrix is the one containing the identity: N itself. And indeed, if Y(gN)=I, then X(g)=I, and so g\in N and gN=N in the first place.

Next, Y is irreducible if and only if X is. We’ll prove this one by using the properties we proved a few days ago. In particular, we’ll calculate the inner product of the character of Y with itself. Writing \psi for the character of Y and \chi for that of X, we find that

\displaystyle\psi(gN)=\mathrm{Tr}(Y(gN))=\mathrm{Tr}(X(g))=\chi(g)

and we use this to calculate:

\displaystyle\begin{aligned}\langle\psi,\psi\rangle&=\frac{1}{\lvert G/N\rvert}\sum\limits_{gN\in G/N}\overline{\psi(gN)}\psi(gN)\\&=\frac{\lvert N\rvert}{\lvert G\rvert}\sum\limits_{gN\in G/N}\overline{\chi(g)}\chi(g)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{gN\in G/N}\sum\limits_{n\in N}\overline{\chi(gn)}\chi(gn)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\overline{\chi(g)}\chi(g)\\&=\langle\chi,\chi\rangle\end{aligned}

Essentially, the idea is that each group element in the kernel N contributes the same to the sum, and this is exactly compensated for by the difference between the sizes of G and G/N. Since the two inner products are the same, either both are 1 or neither one is, and so either both representations are irreducible or neither one is.

We can also run this process in reverse: let G be a finite group, let N be any normal subgroup so we have the quotient group G/N, and let Y be any representation of G/N. We will use this to define a representation of the original group G by “lifting” the representation Y.

So, the obvious choice is to define X(g)=Y(gN). This time there’s no question that X is well-defined, since here we start with a group element and find its coset, rather than starting with a coset and picking a representative element. And indeed, this is easily verified to be a representation.

If Y is faithful, then the kernel of X is exactly N. Indeed, if X(g)=I, then Y(gN)=I. If Y is faithful, then we must have gN=N, and thus g\in N.

And finally, X is irreducible if and only if Y is. The proof runs exactly as it did before.

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October 29, 2010 - Posted by | Algebra, Group theory, Representation Theory

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